GenChem 1111: Mid-Term Study Guide

GenChem 1111: Mid-Term Study Guide

Exp 1: Basic Lab Operation and Dry Lab

  1. Bunsen Burner Operation

    • Valves and Ignition Order: Proper sequence for opening valves and igniting the burner is crucial to ensure safety and functionality.

    • Air-Control Valve:

      • Description: Adjusts the amount of air entering the burner.

      • Color Changes:

      • Sufficient Air: Blue flame.

      • Insufficient Air: Yellow/orange flame (cool flame).

    • Flame Types:

      • Cool Flame: Low temperature.

      • Hot Flame: High temperature used for various reactions.

  2. Density Calculations

    • Formula: D=MVD = \frac{M}{V} (Density = Mass/Volume).

    • Unknown Solid:

      • How to measure the volume of an unknown solid is required for density calculation.

      • Effect of Partial Immersion:

      • If the solid does not immerse completely into water, the recorded volume will be lower than the actual volume.

      • This leads to:

        • D=MVD = \frac{M}{V}

        • If VV is measured lower, then DD will be overestimated (higher than actual density).

    • Unknown Liquid: Calculation of density and the effect of evaporation must be considered.

  3. Reading Volume: Skills for accurately measuring volume using a graduated cylinder.

  4. Significant Figures: Importance of following significant figure rules during calculations.

Exp 2: Empirical Formulas

  1. Magnesium and Oxygen Reaction

    • Combination Reaction: Characterized by two substances (reactants) combining to form one product.

      • Example Reaction: 2Mg+O2→2MgO2Mg + O_2 \rightarrow 2MgO

    • Molar Masses:

      • Magnesium (Mg): 24.32 g/mol.

      • Oxygen (O): 16 g/mol.

    • Calculations: Determine the mass of MgOMgO or O2O_2 reacted from given mass of MgMg.

  2. Crucible Usage: Necessary techniques for safely heating magnesium.

  3. Dry Agent: Identify agents that change color during reactions, indicating progress or completeness.

  4. Concept of Mole:

    • Understand calculations for empirical formulas and mass percentages from experimental data.

    • Balancing Equations: Required for successful calculations in lab reports.

Exp 3: Limiting Reactants

  1. Double Displacement Reaction: Example given:

    • Sr(NO<em>3)</em>2(aq)+2KIO<em>3(aq)→Sr(IO</em>3)<em>2(s)+2KNO</em>3(aq)Sr(NO<em>3)</em>2(aq) + 2KIO<em>3(aq) \rightarrow Sr(IO</em>3)<em>2(s) + 2KNO</em>3(aq)

  2. Limiting Reactants: Definition and importance in reactions; identifies which reactant will determine the maximum amount of product produced.

  3. Yield Percentage Calculations:

    • Definitions:

      • Theoretical Yield: The calculated max product from complete reaction.

    • Practice Example:

      • For reactions of 1.0 mole Sr(NO<em>3)</em>2Sr(NO<em>3)</em>2 with 1.5 moles KIO3KIO_3:

      • Determine grams of precipitate produced and yield percentage if an actual yield of 280g is achieved.

      • Identification of limiting reactant is essential (1.5 moles KIO3KIO_3 is limiting).

      • Calculation of Sr(IO<em>3)</em>2Sr(IO<em>3)</em>2:

        • Reaction Ratio: 1 mol Sr(NO<em>3)</em>2Sr(NO<em>3)</em>2 reacts with 2 moles KIO3KIO_3

        • Calculation:

        • xx moles from limiting reactant gives 0.75 moles

        • Mass Calculation:

          • 0.75extmol∗437.4extg/mol=328.1extg0.75 ext{ mol} * 437.4 ext{ g/mol} = 328.1 ext{ g}.

        • Yield Percentage:

          • 280328.1∗100%=85.3%\frac{280}{328.1} * 100\% = 85.3\%.

  4. Example for Li/N2 Reaction:

    • Reaction: 6Li(s)+N<em>2(g)→2Li</em>3N(s)6Li(s) + N<em>2(g) \rightarrow 2Li</em>3N(s)

    • Molar ratios given align with initial amounts, confirming limiting reactant and resulting product quantity (1 mol of Li3NLi_3N).

    • Example Mixture:

      • 0.810 g of BaCl<em>2⋅2H</em>2OBaCl<em>2\cdot2H</em>2O and Na<em>2SO</em>4Na<em>2SO</em>4 leads to 0.466 g of BaSO4BaSO_4.

    • Calculate mass percentages of components in the original mixture based on theoretical calculations.

      • Ensure to define limiting reactants as necessary in calculations.

Exp 4: Chemistry of Copper

  1. Copper (Cu) Properties: Awareness of physical and chemical characteristics crucial in studies.

    • Five reactions observed:

      • Transformation: Cu→Cu(NO<em>3)</em>2→Cu(OH)<em>2→CuO→CuSO</em>4→CuCu \rightarrow Cu(NO<em>3)</em>2 \rightarrow Cu(OH)<em>2 \rightarrow CuO \rightarrow CuSO</em>4 \rightarrow Cu.

    • Observations regarding color change and state changes important for documenting experiments.

  2. Types of Reactions:

    • Definitions and examples of:

      • Double Displacement Reaction.

      • Decomposition Reaction.

      • Oxidation-Reduction (Redox) Reaction.

  3. Recovery Percent Calculation.

Exp 5: Chemical Reactions

  1. Types of Reactions: Review and understand different reaction types, including:

    • Combination, Decomposition, Single/Double Displacement Reactions.

  2. Redox Reactions: Definitions and process understanding:

    • Oxidation Process: Loss of electrons by a substance.

    • Reduction Process: Gain of electrons by a substance.

    • Identification of reducing agent (e.g., Zinc (Zn) as reducing agent) and oxidizing agent (e.g., Copper (Cu) as oxidizing agent).

  3. Equations: Know how to write different types of chemical equations:

    • Molecular Equation Example:

      • CaCl<em>2⋅2H</em>2O(aq)+K<em>2C</em>2O<em>4⋅H</em>2O(aq)→CaC<em>2O</em>4⋅H<em>2O(s)+2KCl(aq)+2H</em>2O(l)CaCl<em>2\cdot2H</em>2O(aq) + K<em>2C</em>2O<em>4\cdot H</em>2O(aq)\rightarrow CaC<em>2O</em>4\cdot H<em>2O(s) + 2KCl(aq) + 2H</em>2O(l)

    • Complete Ionic Equation:

      • Ca2+(aq)+2Cl−(aq)+2H<em>2O(l)+2K+(aq)+C</em>2O<em>42−(aq)+H</em>2O(l)→CaC<em>2O</em>4⋅H<em>2O(s)+2K+(aq)+2Cl−(aq)+2H</em>2O(l)Ca^{2+}(aq) + 2Cl^{-}(aq) + 2H<em>2O(l) + 2K^{+}(aq) + C</em>2O<em>4^{2-}(aq) + H</em>2O(l)\rightarrow CaC<em>2O</em>4\cdot H<em>2O(s) + 2K^{+}(aq) + 2Cl^{-}(aq) + 2H</em>2O(l)

    • Net Ionic Equation: Result after removing spectator ions:

      • Ca2+(aq)+C<em>2O</em>42−(aq)+H<em>2O(l)→CaC</em>2O<em>4⋅H</em>2O(s)Ca^{2+}(aq) + C<em>2O</em>4^{2-}(aq) + H<em>2O(l)\rightarrow CaC</em>2O<em>4\cdot H</em>2O(s)